Amplitude Truncation and Lp Tail Estimates

Clipping the Amplitude

An $L^p$ function can be unbounded, even though its $p$th power is integrable. A useful way to control its size without discarding its large values is to clip its amplitude: values above $M$ are replaced by $M$, and values below $-M$ by $-M$. The key question is whether this alteration becomes small in $L^p$ norm as $M$ grows. Its answer is also a precise estimate of the contribution from the part of the function whose amplitude exceeds $M$.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; it need not be finite, complete, or sigma-finite. Fix $1\le p<\infty$, and let $f:X\to\mathbb{R}$ be measurable with $\int_X |f|^p\,d\mu<\infty$. Thus $f$ represents an element of the real space $L^p(\mu)$, whose norm is $\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}$. For $M>0$, define the clipping map $T_M:\mathbb{R}\to[-M,M]$ and clipped function $f_M$ by

T_M(t)=\max(-M,\min(t,M)),\qquad f_M(x)=T_M(f(x)).

The function $f_M$ is measurable because $T_M$ is continuous and $f$ is measurable. It is bounded by $M$, though it need not belong to $L^p(\mu)$ if $\mu(X)=\infty$. The difference $f-f_M$, however, is in $L^p(\mu)$: pointwise its absolute value is at most $|f|$.

Pointwise error identity: For every $x\in X$ and $M>0$, $|f(x)-f_M(x)|=(|f(x)|-M)_+$, where $a_+=\max(a,0)$. In particular, $|f-f_M|^p=(|f|-M)_+^p\le |f|^p$.

Indeed, when $|f(x)|\le M$, clipping does not change the value, so both sides of the identity are zero. When $f(x)>M$, the difference is $f(x)-M=|f(x)|-M$; when $f(x)<-M$, its absolute value is $-M-f(x)=|f(x)|-M$. These cases also show why the error is not simply the original function on the set $\{|f|>M\}$: clipping subtracts the level $M$ from its amplitude.

The Tail Estimate and Norm Convergence

Theorem: Let $(X,\mathcal{M},\mu)$ be any measure space, let $1\le p<\infty$, and let $f:X\to\mathbb{R}$ be measurable with $\int_X|f|^p\,d\mu<\infty$. Then, for every $M>0$, $$ \|f-f_M\|_p^p=\int_X(|f|-M)_+^p\,d\mu \le \int_{\{|f|>M\}}|f|^p\,d\mu, $$ and $\lim_{M\to\infty}\|f-f_M\|_p=0$.

Proof. The pointwise identity gives the equality. The integrand $(|f|-M)_+^p$ is zero when $|f|\le M$; on the remaining set, $0\le |f|-M\le |f|$. Integrating this pointwise inequality proves the displayed tail bound. In particular, the error has finite $p$th-power integral.

For convergence, first take positive integers $n$ and put $g_n=(|f|-n)_+^p$. For each $x$, $g_n(x)\to0$, since the real number $|f(x)|$ is finite and eventually at most $n$. Also $0\le g_n\le |f|^p$, and $|f|^p$ is integrable by hypothesis. The Dominated Convergence Theorem therefore applies: the functions $g_n$ are measurable, converge pointwise to zero, and are dominated by the integrable function $|f|^p$. It yields $\int_Xg_n\,d\mu\to0$. Given $\varepsilon>0$, choose an integer $n$ such that $\int_Xg_n\,d\mu<\varepsilon^p$. If $M\ge n$, then $(|f|-M)_+^p\le g_n$ pointwise, so $\|f-f_M\|_p^p<\varepsilon^p$, and hence $\|f-f_M\|_p<\varepsilon$. This proves the limit for all real $M\to\infty$, not only integer levels. $\square$

The assumption $p<\infty$ matters: this argument controls an integral of the error, whereas the $L^\infty$ norm records the essential supremum of the error. The argument also does not require $\mu(X)<\infty$. In fact, although $f_M$ may fail to be in $L^p(\mu)$ on an infinite-measure space, the difference $f-f_M$ is in $L^p(\mu)$ and its norm is well defined.

Worked Example: A Power Singularity on (0,1)

Take $X=(0,1)$ with Lebesgue measure $m$, $p=2$, and $f(x)=x^{-1/4}$. First check the integrability condition:

\int_0^1|f(x)|^2\,dm(x)=\int_0^1x^{-1/2}\,dx=2<\infty.

For $M\ge1$, the condition $f(x)>M$ is equivalent to $x<M^{-4}$, because $x^{-1/4}>M$ if and only if $x<M^{-4}$. The clipping error is therefore supported on $(0,M^{-4})$, and its squared norm is

\begin{aligned} \|f-f_M\|_2^2 &=\int_0^{M^{-4}}(x^{-1/4}-M)^2\,dx\\ &=\int_0^{M^{-4}}\left(x^{-1/2}-2Mx^{-1/4}+M^2\right)\,dx\\ &=2M^{-2}-2M\left(\frac{4}{3}M^{-3}\right)+M^2M^{-4}\\ &=2M^{-2}-\frac{8}{3}M^{-2}+M^{-2} =\frac{1}{3}M^{-2}. \end{aligned}

Thus $\|f-f_M\|_2=1/(\sqrt{3}M)$, which tends to zero. For example, at $M=2$, the squared error is $1/12$, and the norm error is $1/(2\sqrt{3})$. Check: the support has measure $2^{-4}=1/16$; expanding the integrand on that interval gives the three contributions $2M^{-2}=1/2$, $-\tfrac83M^{-2}=-2/3$, and $M^2\cdot\tfrac1{16}=1/4$, whose sum is $\tfrac{6-8+3}{12}=1/12$, as the formula predicts.

Exact Errors for Exponential Tails

A second useful calculation arises when the measure itself has an exponentially decaying density. It illustrates that an unbounded function can have a rapidly decreasing clipping error, and that the exact rate can be computed rather than inferred only from the convergence theorem.

Worked Example: Clipping an Exponential Under a Weighted Measure

Let $X=[0,\infty)$, let $\mathcal{M}$ be the Borel sigma-algebra, and define $\mu(E)=\int_E e^{-2x}\,dm(x)$. This is a finite measure, since $\mu(X)=1/2$. Set $p=1$ and $f(x)=e^x$. The function is integrable because

\int_X|f|\,d\mu=\int_0^\infty e^x e^{-2x}\,dx=\int_0^\infty e^{-x}\,dx=1.

For $M\ge1$, the inequality $e^x>M$ is equivalent to $x>\ln M$. Using the pointwise error identity and integrating with respect to $\mu$,

\begin{aligned} \|f-f_M\|_1 &=\int_{\ln M}^{\infty}(e^x-M)e^{-2x}\,dx\\ &=\int_{\ln M}^{\infty}e^{-x}\,dx-M\int_{\ln M}^{\infty}e^{-2x}\,dx\\ &=\frac{1}{M}-M\left(\frac{1}{2M^2}\right) =\frac{1}{2M}. \end{aligned}

For instance, $M=2$ gives error $1/4$. Check: the first integral in the second line is $e^{-\ln 2}=1/2$, and the subtracted term is $2\cdot e^{-2\ln2}/2=1/4$; their difference is $1/4$. The exact formula also tends to zero, in agreement with the theorem.

These computations are instances of the same principle: the error is determined by the values above the clipping level, not by the values below it. A useful general bound, already contained in the theorem, is $\|f-f_M\|_p^p\le\int_{\{|f|>M\}}|f|^p\,d\mu$. It may be less sharp than an exact calculation, but it directly relates the clipping error to the $p$th-power mass in the high-amplitude tail.

Worked Example: A Discrete Power Tail

Let $X=\mathbb{N}=\{1,2,\ldots\}$, take counting measure $\mu$, $p=2$, and $f(n)=1/n$. Then $f\in L^2(\mu)$, since $\sum_{n=1}^\infty n^{-2}<\infty$. For every $M\ge1$, $0\le f(n)\le1\le M$ for all $n$. Clipping does nothing: $f_M(n)=f(n)$, and therefore $\|f-f_M\|_2^2=\sum_{n=1}^\infty0=0$. Check: the tail set $\{n:|f(n)|>M\}$ is empty, so the tail bound also gives zero. This example emphasizes that a function can have an infinite domain and infinitely many nonzero values while having no clipping error at a given level.

A Common Mistake: Clipping Is Not Cutting Off

Amplitude clipping must be distinguished from cutting off the tail, which replaces values above $M$ by zero. For clipping, the difference at a positive value $f(x)>M$ is $f(x)-M$; for cutting off, the difference is $f(x)$. Confusing these operations gives an incorrect error formula.

Worked Example: Checking the Error on a One-Point Space

Let $X=\{x_0\}$, let $\mathcal{M}$ contain all subsets, give the point measure $1$, and take $p=2$, $f(x_0)=2$, and $M=1$. Then $f_M(x_0)=1$, so the actual error is

\|f-f_M\|_2^2=|2-1|^2=1.

The incorrect formula $\int_{\{|f|>M\}}|f|^2\,d\mu$ would give $2^2=4$. That quantity is the squared error for replacing $f$ by zero on the tail, not for clipping it to height $M$. The correct integrand is $(|f|-M)_+^2$, which gives $(2-1)^2=1$. Check: the tail bound is valid here and reads $1\le4$, while the exact identity gives the actual error $1$.

The theorem is useful whenever an argument first needs bounded functions but ultimately concerns an $L^p$ function: clipping supplies bounded approximants while quantifying exactly how much is changed. It controls amplitude only. It does not restrict the function to a finite-measure set, nor does it assert that the clipped functions have finite $L^p$ norm on an infinite-measure space. Those are separate issues.

Check Your Understanding

For each question, identify the measure space and verify the integrability hypothesis before using the tail estimate.

For $f(x)=x^{-1/3}$ on $(0,1)$ with Lebesgue measure and $p=2$, determine whether $f\in L^2$. What feature of the theorem explains your answer?
For $f(x)=x^{-1/4}$ on $(0,1)$ with $p=2$, find the set on which clipping at level $M\ge1$ changes the function.
Explain why $\|f-f_M\|_p^p$ is bounded by $\int_{\{|f|>M\}}|f|^p\,d\mu$, and state when equality in this bound holds pointwise on the tail.
On a one-point space of measure $1$, let $f=5$, $p=1$, and $M=3$. Compute the clipping error and the error from replacing $f$ by zero.
Why does boundedness of $f_M$ alone not imply $f_M\in L^p(\mu)$ when the measure space has infinite measure?